# Can a minimum variance unbiased estimator be inconsistent?

Title says it all. I'm guessing it shouldn't be possible for an estimator to be UMVUE and yet not consistent. Is there a counter-example to this? Or is there a proof that it can't happen?

One counterexample is when there's no consistent estimator. Suppose $$X_i\sim N(\mu_i,1)$$ where $$\mu_i$$ are all distinct unknown parameters. The UMVUE of any $$\mu_i$$ is $$X_i$$, but it's not consistent.
You can make matters worse. Suppose you a have single $$X\sim N(\theta,1)$$ together with $$Y_i\sim\text{Bernoulli}(p)$$ with $$\mathrm{logit}\,p=\theta$$ for $$i=1,\dots,n$$. Then $$\mathrm{logit}\, \bar Y_n$$ is a consistent estimator for $$\theta$$ as $$n\to\infty$$, but no function of $$Y_i$$ is unbiased for $$\theta$$. This means the UMVUE of $$\theta$$ is just $$X$$, which is not consistent.
However, for iid $$X_i$$, $$i=1,\dots,n$$, if we assume there exists some unbiased estimator $$\tilde\theta$$ that has finite variance for all $$n$$ greater than some $$n_0$$, the MVUE must be consistent. We can divide the $$n$$ observations into $$[n/n_0]$$ blocks of size at least $$n_0$$ and take the average of $$\tilde\theta$$ for each block. If $$\sigma^2$$ is the variance of each block-specific $$\tilde\theta$$ then the variance of the average is $$\sigma^2/[n/n_0]$$, which goes to zero as $$n\to\infty$$. So there is a consistent unbiased estimator and so the MVUE is also consistent.